Introduction Sodium hydroxide is hygroscopic and absorbs water from the air when you place it on the balance for massing. I doubt it could be 50% - 50g/100ml would be 500g/L - this would be about 12-13M - it is very unlikely that you would be provided a 50% solution as this would be very corrosive and is not suitable as a standard as it readily absorbs CO2. How do I find the percentage of KHP in a sample with the following information: Weighed out a certain amount of KHP (MW=204.22g/mol) into each Erlenmeyer flask: Added 50mL of distilled water to each of the flask. Using flask 1 details, moles NaOH = 0.0247 x 0.0625 = 1.54375 x 10^-3 If so is it a 1:1? 0.0625M? Then the molarity was determined from this titration and the value used to determine the percentage composition of KHP in another experiment. find the percentage of KHP in a sample by titration with NaOH, (M NaOH)(mL titrated) / 1000mL = mol NaOH, finding Ksp for a solution where solubility is known, Calculate the concentration of ion in the given solution using Ksp, determining if a reaction forms a precipitate when two solutions are mixed, Calculate [Ag+] in this saturated solution of Ag2SO4, Find the solubility product when solubility is given, calculate the solubilitry constant (Ksp) for cupric hydroxide Cu(OH)2, Ksp solubility product constant at 25C for AgCl, Find the molar concentration of F- ions if given Ksp for CaF2 and [Ca2+] ionsca. Question: Question 9 2 Pts KHP (aq) + NaOH(aq) — NaKHP (aq) + H2O (1) Trial V: (mL) V+(mL) 47.25 1 35.81 2 3 A Student Masses Out 1.033 Grams Of KHP, Which Is Dissolved In 25.00 ML Of DI Water. If the molarity of the NaOH was 0.0625M, this is what must be used in the calculations. That was part1 of the experiment whereas part2 was the unknown KHP and the one I wrote about. In fact when I taught, we used KHP as a primary standard and determined the molarity of NaOH (which then became a secondary standard). The NaOH solution is standardized using the titration of a primary standard of KHP (Figure 2). Is it possible to even get that high of the percentage of KHP in a sample? What is the exact molarity of the NaOH solution? By: Juno Kim Introduction In this experiment the concentration of potassium hydrogen phthalate (KHP) in an unknown sample was determined through volumetric analysis. That's how I got my NaOH solution. It reacts with a base to form a salt, C6H4 COO-Na+ COO-K+. This is minimized by using a small amount of a dilute solution of the indicator and by choosing an indicator that changes color close to the equivalence point pH. (2.46e-3)(204.224) / .4924 x 100 = 102.19%, Flask#3: However, it still does not explain how the NaOH used was 0.0625M. With a standard deviation being 2.41 the experiment concluded with 95% confidence that the KHP concentration in the unknown sample measures between 25.58% and 30.4%. Mole KHP x Formula mass will give the mass of KHP present in the flask. And I saw my professor doing the molarity when he was checking my other work, but I don't remember how he did it lol. You will need to find the missing details to show that the molarity was 0.0625M. Because the In-1 and Hin have different colors we can see this change. HC8H4O4-1(aq) + OH-(aq) → C8H4O4-2 (aq) + H2O(l). The volumetric analysis allows for the calculation of the unknown amount or concentration by a reaction with a known amount of reagent. From the information you have given, it appears you were given the molarity of NaOH to allow you to calculate the % composition of KHP, I can't say anything about the concentration of NaOH. When KHP and NaOH combine, a positive hydrogen ion leaves the KHC8H4O4 and a negative hydrogen atom leaves the NaOH. You will use this primary standard to determine the concentration of a sodium hydroxide solution. Dividing this by the sample mass and multiplying by 100 will give the percentage KHP in the sample. Do I need to form a relationship between the KHP and NaOH? We are doing titration of KHP and NaOH in lab tomorrow, and in our manual, they give the following equation: HKC 8 H 4 O 4 + NaOH --->NaKC 8 H 4 O 4 +H 2 O. NaOH +KHP -----> H2O + KNaP. What is the exact molarity of the NaOH solution? Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. BUT since that's the case, do I find the new molarity of NaOH using my unknown-KHP masses? This can then be compared to the NaOH to find the number of moles NaOH has.. Calculate The Molarity Of The NaOH Solution. An indicator is added to signal the endpoint has been reached. therefore number of moles of KHP is 1.0874/(39.1+1+31) =0.015mol. The titrations yielded an average concentration of 27.99% KHP in unknown sample. Since the indicator reacts with some of the titrant and the indicator may not change at the exact pH of the equivalence point, a small error in introduced in the titration. A buret filled with the titrant (NaOH solution) is used to measure the volume of NaOH solution added to the known amount of acid in a flask. Sodium hydroxide is not a primary standard and would need to be standardised prior to use. An acid-base titration is a procedure used to compare the amount (moles) of acid in one sample with the amount (moles) of base in another. If the molarity of the NaOH was 0.0625M, this is what must be used in the calculations. Average the three value from the 3 titrations will give the answer. This water will prevent you from being able to find the exact mass of sodium hydroxide. When KHP and a base a reacted, a neutralization reaction occurs that is represented by the following equation: When acid is added to the indicator the equilibrium shifts to the left. ( Log Out /  References. Change ), You are commenting using your Google account. (1.91e-3)(204.224) / .3907 x 100 = 99.88%, Then take average of all three to rfind the final %KHP: ( Log Out /  We are doing titration of KHP and NaOH in lab tomorrow, and in our manual, they give the following equation: HKC 8 H 4 O 4 + NaOH --->NaKC 8 H 4 O 4 +H 2 O.

khp + naoh

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